Two short bar magnets of length $1cm$ each have magnetic moments $1.20 \mathrm{A} {\mathrm{m}}^2$ and $1.00 \mathrm{A} {\mathrm{m}}^2$ respectively. They are placed on a horizontal table parallel to each other with their $\text{N}$ poles pointing towards the south. They have a common magnetic equator and are separated by a distance of $20.0 \mathrm{cm}$. The value of the resultant horizontal magnetic induction at the mid-point $\text{O}$ of the line joining their centres is close to
( Horizontal component of earth's magnetic induction is $3.6\times {10}^{-5} \mathrm{Wb} {\mathrm{m}}^{-2}$ )

• A

  $3.50\times {10}^{-4} \mathrm{Wb} {\mathrm{m}}^{-2}$
• B

  $5.80\times {10}^{-4} \mathrm{Wb} {\mathrm{m}}^{-2}$
• C

  $3.6\times {10}^{-5} \mathrm{Wb} {\mathrm{m}}^{-2}$
• Correct answer
  D

  $2.56\times {10}^{-4} \mathrm{Wb} {\mathrm{m}}^{-2}$

Solution

${\text{B}}_{\text{net}}={\text{B}}_1+{\text{B}}_2+{\text{B}}_{\mathrm{H}}$

$=\frac{{\mu }_0}{4\pi }\frac{{\text{M}}_1}{{\text{x}}^3}+\frac{{\mu }_0}{4\pi }\frac{{\text{M}}_2}{{\text{x}}^3}+{\text{B}}_{\text{H}}$

$=\frac{\left\{{10}^{-7}\left(1.2+1\right)\right\}}{{\left(0.1\right)}^3}+3.6\times {10}^{-5}$

$=\text{2.56}\times {\text{10}}^{-4}\text{ Wb}/{\text{m}}^2$

Hence, the correct answer is option: 'D'.